Question

questões de racionalização e expressões com radicais, alguém pode me ajudar?​

Answer 1

Resposta:

Questão 16:

a)

$( \sqrt{18} + \sqrt{98} + \sqrt{200}):(2 .\sqrt{2} +\sqrt{8} )$

$( \sqrt{9.2} + \sqrt{49.2} + \sqrt{100.2}):(2 .\sqrt{2} +\sqrt{4.2} )$

$( 3\sqrt{2} + 7\sqrt{2} + 10\sqrt{2}):(2 .\sqrt{2} +2\sqrt{2} )$

$( 20\sqrt{2}):(4\sqrt{2})$

$= 5$

b)

$( 10.\sqrt{27}+10\sqrt{3} ):(10\sqrt{3})$

$( 10.\sqrt{9.3}+10\sqrt{3} ):(10\sqrt{3})$

$( 10.3.\sqrt{3}+10\sqrt{3} ):(10\sqrt{3})$

$( 30.\sqrt{3}+10\sqrt{3} ):(10\sqrt{3})$

$( 40\sqrt{3} ):(10\sqrt{3})$

$=4$

Questão 17:

a) $\frac{6}{\sqrt{3} } =\frac{6.\sqrt{3} }{\sqrt{3} .\sqrt{3} } =\frac{6\sqrt{3} }{3} =2\sqrt{3}$

b) $\frac{2}{3\sqrt{5} } =\frac{2\sqrt{5} }{3\sqrt{5}.\sqrt{5} } =\frac{2\sqrt{5} }{3.5} =\frac{2\sqrt{5} }{15}$

c) $\frac{4}{\sqrt[8]{2^5} } =\frac{4.\sqrt[8]{2^5} }{\sqrt[8]{2^5}.\sqrt[8]{2^5}} =\frac{2^2.2^{5/8}}{2^{10/8}}=\frac{2^{21/8}}{2^{10/8}}= 2^{\frac{21-10}{8} }=2^{11/8}=\sqrt[8]{2^{11} }=2.\sqrt[8]{2^3} = 2.\sqrt[8]{8}$

d) $\frac{3}{4+\sqrt{5} } =\frac{3 . (4-\sqrt{5}) }{(4+\sqrt{5}) . (4-\sqrt{5}) } = \frac{12 - 3\sqrt{5} }{4^2 - 5} =\frac{12-3\sqrt{5} }{16-5}=\frac{12-3\sqrt{5} }{11}$

e) $\frac{\sqrt{2} }{\sqrt{10} +\sqrt{3} } = \frac{\sqrt{2}.(\sqrt{10} -\sqrt{3}) }{(\sqrt{10} +\sqrt{3}).(\sqrt{10} -\sqrt{3}) } = \frac{\sqrt{20}-\sqrt{6} }{10-3} =\frac{\sqrt{20}-\sqrt{6}}{7}$

f) $\frac{\sqrt{3}+1 }{\sqrt{3}-1 } = \frac{(\sqrt{3}+1).(\sqrt{3}+1) }{(\sqrt{3}-1).(\sqrt{3}+1) }=\frac{3 + 2\sqrt{3}+1 }{3 - 1} =\frac{4+2.\sqrt{3} }{2} =\frac{2.(2+\sqrt{3}) }{2}=2+\sqrt{3}$